Questions over a specific case of the Muntz-Szasz theorem proof On page 157 of this site:
http://arxiv.org/pdf/0710.3570.pdf
the author is proving a specific case of one direction of the Muntz-Szasz theorem.
I do not understand the following 3 claims: 
1) For $\lambda > 0$ and $x \in (0,1)$, $\lambda x^{\lambda}(1-x) <1$. 
Why? I have tried various methods, including induction, and got nowhere.
2) The above claim implies (apparently) for some fixed positive integer $q$:
$||Q_n||_{C[0,1]} \leq |1-{q \over \lambda_{n}}|||Q_{n-1}||_{C[0,1]}$. 
See the article for the definition of $Q_n$. It's on page 156 at the beginning of the section.
Why does this follow from my first claim?
3) Why is the last line of the proof true? 
Specifically, as $\lambda_n$ goes to $\infty$ by assumption in the proof, how does the last product go to $0$?
I'd suggest quickly reading the section if you're confused by my question. It's very simple up to this point.
Thanks!
 A: This answers #1.
Fix $x \in (0,1)$ and set $f(\lambda) = \ln(\lambda x^\lambda (1-x)) = \ln \lambda + \lambda \ln x + \ln(1-x)$.  Then $f'(\lambda) = \frac{1}{\lambda} + \ln x$, so if $\lambda_0 = -\frac{1}{\ln x}$, we have that $f$ is increasing on $(0, \lambda_0)$, decreasing on $(\lambda_0, \infty)$, and so achieves its global maximum at $\lambda = \lambda_0$.
Now $f(\lambda_0) = -\ln(-\ln x) - 1 + \ln(1-x)$.  But by the elementary inequality $-\ln x > 1-x$ for $x \in (0,1)$, and the fact that $\ln$ is an increasing function, we have $\ln(-\ln x) > \ln(1-x)$.  Hence $f(\lambda_0) < -1$.
We conclude that $f < -1$, which is to say
$$\lambda x^\lambda (1-x) < \frac{1}{e} < 1.$$
A: Nate's answer is correct but personally I'm not a big fan of the "start with the answer and check derivatives" approach to proving bounds, because it conveys no insight into how they were originally obtained. (Granted, this is less of an issue when one side of the bound is a constant.)
Often there is some integral lurking in the background. Here, we can write 
$$
\begin{align*}
\lambda x^\lambda(1-x) 
&= \lambda \int_0^x t^{\lambda - 1}\left[\lambda-(\lambda+1)t\right]\,\mathrm{d}t \\
&\le \lambda \int_0^{\frac{\lambda}{\lambda+1}} t^{\lambda - 1}\left[\lambda-(\lambda+1)t\right]\,\mathrm{d}t \\
&= \lambda^2\left(\frac{\lambda}{\lambda+1}\right)^\lambda \int_0^1 u^{\lambda - 1}(1-u)\,\mathrm{d}u \\
&= \left(\frac{\lambda}{\lambda+1}\right)^{\lambda+1} \\ &<1.
\end{align*}$$
